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Question

Mathematics Question on Trigonometric Identities

A vertical lamp-post 6 m height stands at a distance of 2 m from a wall, 4 m high. A 1 -5 m tall man starts to walk away from the wall on the other side of the wall in line with the lamp-post. The maximum distance to which the man can walk, remains in the shadow is

A

52m\frac {5} {2} m

B

32m\frac {3} {2} m

C

4 m

D

none of these.

Answer

52m\frac {5} {2} m

Explanation

Solution

BD is the reqd. distance From similar triangles 64=QRBRi.e.,32=2+BRBRBR=4\frac{6}{4} = \frac{QR}{BR} \, i.e., \, \frac{3}{2} = \frac{2 + BR}{BR} \Rightarrow BR = 4 43/2=4DRDR=32\frac{4}{3/2} = \frac{4}{DR} \, \therefore \, DR = \frac{3}{2} BD=BRDR=432=52\therefore \, BD = BR - DR = 4 - \frac{3}{2} = \frac{5}{2}